Light OJ 1026

题目大意：双向联通图， 现在求减少任意一边使图的联通性改变，按照起点从小到大列出所有这样的边

解题思路：双向边模版题 tarjan算法

代码如下：

#include<bits/stdc++.h>
using namespace std;

const int N = 100003;
vector<int>vec[N];
pair<int, int>edge[N];
int dfn[N], low[N];
int res, ans;

void tarjan(int u, int f)
{
    dfn[u] = low[u] = res ++;
    for(int i = 0; i < vec[u].size(); ++ i)
    {
        int v = vec[u][i];
        if(dfn[v] == -1)
        {
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
        }
        else if(f != v)
            low[u] = min(low[u], dfn[v]);

        if(dfn[u] < low[v])
        {
            if(u > v)
                edge[ans ++] = make_pair(v, u);
            else
                edge[ans ++] = make_pair(u, v);
        }
    }
}

void solve(int cases)
{
    for(int i = 0; i < N; ++ i)
        vec[i].clear();

    int n;
    scanf("%d", &n);

    for(int i = 0; i < n; ++ i)
    {
        int a, b, c;
        scanf("%d (%d)", &a, &b);
        for(int j = 1; j <= b; ++ j)
        {
            scanf("%d", &c);
            vec[a].push_back(c);
        }
    }

    memset(dfn, -1, sizeof(dfn));
    memset(low, -1, sizeof(low));

    ans = res = 0;
    for(int i = 0; i < n; ++ i)
    {
        if(dfn[i] == -1)
            tarjan(i, -1);
    }
    sort(edge, edge+ans);
    printf("Case %d:
", cases);
    printf("%d critical links
", ans);
    for(int i=0; i<ans; ++ i)
        printf("%d - %d
", edge[i].first, edge[i].second);
}

int main()
{
    int T;
    scanf("%d", &T);
    for(int i = 1; i <= T; ++ i)
        solve(i);
    return 0;
}